Question: Is ${851760}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {851760}= &&{8}\cdot100000+ \\&&{5}\cdot10000+ \\&&{1}\cdot1000+ \\&&{7}\cdot100+ \\&&{6}\cdot10+ \\&&{0}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {851760}= &&{8}(99999+1)+ \\&&{5}(9999+1)+ \\&&{1}(999+1)+ \\&&{7}(99+1)+ \\&&{6}(9+1)+ \\&&{0} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {851760}= &&\gray{8\cdot99999}+ \\&&\gray{5\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{7\cdot99}+ \\&&\gray{6\cdot9}+ \\&& {8}+{5}+{1}+{7}+{6}+{0} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${851760}$ is divisible by $9$ if ${ 8}+{5}+{1}+{7}+{6}+{0}$ is divisible by $9$ Add the digits of ${851760}$ $ {8}+{5}+{1}+{7}+{6}+{0} = {27} $ If ${27}$ is divisible by $9$ , then ${851760}$ must also be divisible by $9$ ${27}$ is divisible by $9$, therefore ${851760}$ must also be divisible by $9$.